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I believe that I have shown that $(2,3)$ is non-principal in $\mathbb{Z}[x]$. My outline goes something like this: Assume that $(2,3) = (f(x))$ then $f(x)$ divides 2 and 3, that is 2 = $f(x)g(x)$ and 3 = $f(x)q(x)$, so the sum of the degrees of $f(x)$ and $g(x)$ is 0 and the same is true for $f(x)$ and $q(x)$. Let $f(x)=c$ where $c$ is a constant. This would then mean that $(2,3) = (\pm 1) = \mathbb{Z}[x]$, a contradiction.

I was wondering whether this argument might work for the case of $\mathbb{Q}[x]$ but it seems that you could form any polynomial from $\mathbb{Q}[x]$ using the generator $(2,3)$.

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To the first paragraph: $3 - 2 = 1$ is contained in the ideal $(3, 2)$ of $\mathbf Z[x]$. So it seems to me that this is the unit ideal in $\mathbf Z[x]$, hence in $\mathbf Q[x]$. –  Dylan Moreland Mar 30 '12 at 23:11
Indeed. Your "contradiction" is no contradiction at all: $(2,3)=(1)$, in $\mathbb{Z}[x]$ as in $\mathbb{Z}$. –  Chris Eagle Mar 30 '12 at 23:12
Oh thanks for noticing that. I overlooked the obvious! –  Low Scores Mar 30 '12 at 23:18
I think you overlooked the obvious because you seem to have wrote "a contradiction" like a way to end a proof. I know it will seem trivial, but when you end a proof by saying "a contradiction", it's because something must have been contradicted ; always remember to know what is. –  Patrick Da Silva Mar 31 '12 at 0:54

3 Answers 3

up vote 8 down vote accepted

Note that $(2,3) = (1)$ in $\mathbb{Z}$, hence $(2,3)=(1)$ in any larger ring (that has the same unity) as well, in particular in $\mathbb{Z}[x]$ and in $\mathbb{Q}[x]$. There is no contradiction, since in fact $(2,3)$ does contain $1$: $(2,3)$ contains the difference of $3$ and $2$, after all.

On the other hand, $(2,x)$ is non-principal in $\mathbb{Z}[x]$.

More generally: if $D$ is a Unique Factorization Domain that is not a field, and $p$ is a prime of $D$, then $(p,x)$ is not principal in $D[x]$.

On the other hand, if $D$ is a field, then every ideal of $D[x]$ is principal, since $D[x]$ is a Euclidean domain.

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$\mathbb{Q}[x]$ is a Euclidean Domain so all ideals $\mathbb{Q}[x]$ are principal.

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Note $\ $ When considering the persistence of coprimality (comaximality) it is essential to consider the persistence of $1$, i.e. that the (inclusion) homomorphism preserves $1$.$\ $ If $\rm\:(r, s) = (1)\:$ in $\rm R$ then this persists in any superring $\rm S \supset R$ that has the same $1$, but it may fail if not, e.g. $\:3-2 = 1$ $\Rightarrow$ $(3,2) = (1)$ does not persist when embedded in $\mathbb Z^2$ via $\rm\:n\to (n,0)$ since in the superring it becomes $(3,0)-(2,0) = (1,0),$ but $(1,0)\ne (1,1) = 1_{\mathbb Z^2}.\:$ So comaximality does not persist in this case. Occasionally even experienced mathematicians make serious errors by overlooking this subtlety.

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Lol. Superrings. =) –  Patrick Da Silva Mar 31 '12 at 0:52
I'm guessing it was a bit too long for a comment; I've added the caveat. –  Arturo Magidin Mar 31 '12 at 2:12
@Arturo Indeed, deja vu... –  Bill Dubuque Mar 31 '12 at 2:24

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